Video Transcript
Determine the integral of the cos
of seven 𝑥 minus seven with respect to 𝑥.
The question is asking us to
evaluate the integral of a composite function. It’s the integral of the
composition of a trigonometric function and a linear function. And there’s a lot of different ways
we could think about evaluating this integral. For example, we might want to try
using our multiple angle formulas to rewrite our integrand. However, this is also the integral
of a composite function. So we should also check if there’s
a suitable 𝑢 substitution we can use to make this integral easier. And in this case, this is true. We’ll set 𝑢 to be our inner
function. That’s the linear function seven 𝑥
minus seven. Then, we can differentiate 𝑢 with
respect to 𝑥. Since this is a linear function, we
just get the coefficient of 𝑥 which is seven.
At this point, we need to be a
little bit careful. Normally, we would check to see if
d𝑢 by d𝑥 appears in our integrand so we could use this to make or integral
simpler. However, on first glance, this
doesn’t seem to be the case. However, if we multiply our
integrand by seven and then multiply our entire integral by one over seven, we don’t
change the value of our integral. But now we have d𝑢 by d𝑥
appearing in our integrand. And in doing this, we’ve rewritten
our integral to be of the form one-seventh times the integral of 𝑔 prime of 𝑥
times 𝑓 of 𝑔 of 𝑥 with respect to 𝑥. We can use the reverse chain
rule.
So let’s try and evaluate our
integral by using the substitution 𝑢 is equal to seven 𝑥 minus seven. We’ve already shown d𝑢 by d𝑥 is
equal to seven. And remember, d𝑢 by d𝑥 is not a
fraction. However, when we’re integrating by
substitution, we can treat it a little bit like a fraction. This gives us the equivalent
statement in terms of differentials d𝑢 is equal to seven d𝑥. We’re now ready to evaluate our
integral by using substitution. First, the integral of the cos of
seven 𝑥 minus seven with respect to 𝑥 is equal to one-seventh times the integral
of seven times the cos of seven 𝑥 minus seven with respect to 𝑥.
Next, we set 𝑢 to be our inner
function seven 𝑥 minus seven. We then show that d𝑢 by d𝑥 is
equal to seven. And this is equivalent to saying
d𝑢 is equal to seven d𝑥. In other words, in our integration
by substitution, we can replace seven d𝑥 with d𝑢. So using integration by
substitution, we get one-seventh multiplied by the integral of the cos of 𝑢 with
respect to 𝑢. And we know how to evaluate this
integral. The integral of the cos of 𝜃 with
respect to 𝜃 is equal to the sin of 𝜃 plus the constant of integration 𝑐. So applying this rule with 𝑢
instead of 𝜃, we get one-seventh times the sin of 𝑢 plus the constant of
integration we’ll call 𝑐 one.
Distributing one-seventh over our
parentheses, we get one-seventh times the sin of 𝑢 plus 𝑐 one over seven. And we’ll simplify our answer. First, 𝑐 one is a constant. So 𝑐 one divided by seven is also
a constant. So we’ll just replace this with a
new constant we’ll call 𝑐. Next, remember, our original
integral was in terms of 𝑥. So we want to give our answer in
terms of 𝑥. We’ll do this by reusing our
substitution 𝑢 is equal to seven 𝑥 minus seven. This gives us one-seventh times the
sin of seven 𝑥 minus seven plus 𝑐. And this is our final answer.
Therefore, by using integration by
substitution, we’ve shown the integral of the cos of seven 𝑥 minus seven with
respect to 𝑥 is equal to one-seventh times the sin of seven 𝑥 minus seven plus our
constant of integration 𝑐.
